Question: $ g(x) = \int_{-1}^{x}(8 - t)\,dt\,$ $ g\,'(1)\, =$
Answer: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = 8 - t$ is continuous on $[-1,1]$. Applying the theorem We're given: $ g(x) = \int_{-1}^{x}(8 - t)\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =8 - x$ Evaluating $g'(1)$ $ g'(1)= 8 - 1 = 7$ The answer: $g'(1)=7$